Iterated backpropagation and total speed

The preceding lesson found one earlier frequency component. We now repeat that operation and prove quantitative bounds for an earlier point at a prescribed physical time scale. The proof keeps the original viscosity, pressure, spatial center, frequency, time interval and amplitude.

The central step adds short, disjoint intervals on which the actual velocity must remain large. Their total contribution is bounded by the full total-speed estimate. This controls the sum of the spatial steps and forces an earlier frequency in an explicit range. The final estimates depend on the critical velocity bound and the original parameters.

Read Frequency backpropagation with the original parameters for the one-step operation and Global nonlinear energy and total speed for the complete total-speed coefficient. The human comparison is Terence Tao, Quantitative bounds for critically bounded solutions to the Navier–Stokes equations, version 2, original author article.tex 284–294 and 614–658. Section 6 proves the exact map to its stated time separation, including an explicit finite choice of the hierarchy exponent. The subsequent weighted heat argument is still needed for the general large critical-velocity endpoint.

1. The original inputs and complete total speed

Retain the original smooth source-class solution

on \([t_0-T,t_0]\times\mathbb R^3\), with \(T>0\). The smooth source class is that of NS-FLUID-12: \(u\) and its spatial derivatives lie in \(L^\infty_tL^2_x\). Sobolev embedding therefore gives a finite uniform spatial maximum, used below only to prove termination. The force is zero, as required by the one-step operation; the forced total-speed estimate alone does not supply a forced backpropagation theorem.

Keep the Fourier convention, \(\phi,p(\xi),P_N\), and all original positive constants of NS-FLUID-15. The pressure \(p(t,x)\) and the Fourier symbol \(p(\xi)\) are distinguished by their domains. In particular \(C_P=\|\mathcal F^{-1}p(\xi)\|_1>0\), \(0<b\leq b_0\), \(0<\eta\leq1/256\), \(d=b\eta^6/D_u>0\), \(D\geq4d\), and \(L>0\). Those constants retain their full dependence on the original \(U,\nu\) and the fixed cutoff.

Define the four actual coefficients from NS-FLUID-12 (8.4)–(8.6):

All symbols on the right are the complete finite kernel and Sobolev constants defined and proved in NS-FLUID-12. In particular the subscripted \(d_r\) are its companion-kernel constants; the unsubscripted \(d\) is the positive time-gap constant of NS-FLUID-15. There is no discarded viscosity factor or unspecified new constant.

For every actual \(0<S\leq T/2\), apply that proved estimate on the earlier interval \([t_0-2S,t_0]\), with heat starting time \(t_0-2S\) and receiving interval \([t_0-S,t_0]\). This is a restriction of the original solution, not a change of coordinates. Its critical norm is at most the original \(U\). The proof of NS-FLUID-12 uses only that upper bound; each coefficient is nonnegative. It gives

Thus the complete total speed is available on every receiving interval below, with the required preceding heat interval inside the original domain.

2. An earlier point with explicit bounds

If \(U=0\), then \(u=0\) and no positive target amplitude occurs. Assume \(U>0\), retain any \(0<b\leq b_0\), and set

Every constant is finite and positive, and the choices are sequential. Directly \(0<\rho\leq1/32\) because \(D\geq4d\) and \(\eta\leq1\). Also \(0<\gamma\leq1/2\), by the second entry in the maximum defining \(\beta\).

Suppose

We prove that an actual dyadic frequency \(N_i\in N_0\,2^{\mathbb Z}\) and an earlier point satisfy

All bounds depend only on the original \(U,\nu,b\) and the fixed cutoff. The auxiliary smooth maximum is absent from (2.1)–(2.2). The original amplitude is retained.

3. Crossing the actual receiving time

The nonempty interval for \(S\) implies

In particular the initial point satisfies the one-step threshold. Iterate NS-FLUID-15 (2.6) while \(t_j\geq t_0-T/2\) and \(N_j\geq2\sqrt{D/T}\). The already proved operation supplies each next point. For every step,

Here all frequencies remain in the single original dyadic lattice.

For completeness, let \(V=\sup_{[t_0-T,t_0]}\|u(t)\|_\infty\). Every selected frequency obeys \(N_j\leq C_PV/b\). The positive amplitude gives \(V>0\), and hence every step decreases time by at least \(d(b/(C_PV))^2>0\). Each step begins at time at least \(t_0-T/2\) and lasts at most \(T/4\). Thus every selected time is at least \(t_0-3T/4\). An infinite chain would contradict these two facts, so it stops at some finite \(n\geq1\). At that point either \(t_n<t_0-T/2\) or \(N_n<2\sqrt{D/T}\).

In the latter case the incoming frequency comparison gives \(N_{n-1}\leq\eta^{-3}N_n<2\eta^{-3}\sqrt{D/T}\). Consequently

In the former case \(t_0-t_n>T/2\geq2\rho T\). Both cases therefore put the stopping point strictly before \(t_0-S\). This is the exact dimensional and ordered inequality needed for the crossing; no reversed comparison is used.

The first step has \(t_0-t_1\leq DN_0^{-2}\leq S/2\). Let \(m\) be the largest index with \(t_m\geq t_0-S\). Strictly decreasing times now give

This proves the complete crossing and the lower sum of squared inverse frequencies in the original time variable.

4. Disjoint intervals control the total displacement

For each \(j=0,\ldots,m-1\), keep the actual fixed point \(x_j\) and define

The derivative estimate proved in NS-FLUID-15 is \(|\partial_tP_{N_j}u|\leq D_uN_j^3\). The fundamental theorem of calculus on \(J_j\) gives

The lower step bound in (3.1) puts each \(J_j\) strictly inside \([t_{j+1},t_j]\) except possibly at its top endpoint. Their interiors are pairwise disjoint, and all are contained in \([t_m,t_0]\subset[t_0-S,t_0]\). In particular the argument uses only intervals where both the derivative estimate and (1.3) have already been proved.

The original projection kernel has scaled \(L^1\) norm \(C_P\). Hence \(\|u(t)\|_\infty\geq bN_j/(2C_P)\) on \(J_j\). Adding the integrals over the disjoint intervals and using (1.3) gives

The last frequency satisfies \(N_m^{-1}\leq\eta^{-3}N_{m-1}^{-1}\). Since \(m\geq1\), it follows that

This replaces the auxiliary smooth-maximum displacement estimate by a bound in the original critical norm, viscosity and amplitude. Every spatial step and every persistence interval has been kept.

Figure 1. The left panel displays a geometric example with \(t_0=1,S=1/4,d=1,D=4\), frequencies \(8,4,4,2\), and steps of length \(2/N_j^2\). The crossing is \(t_2=27/32\geq3/4>23/32=t_3\), so \(m=2\). The two retained persistence intervals have lengths \(1/128\) and \(1/32\) and satisfy exactly the containment in (4.1)–(4.2). Only the three steps up to the crossing are shown. These values illustrate interval geometry; they are not evaluations of the PDE constants in NS-FLUID-15 or (2.1), and are not claimed to meet the theorem's scale-selection hypotheses. The right panel uses the same original frequencies: \(r_0=1/8,r_1=r_2=1/4\). The three exact blue areas total \(9/64\), and the surrounding rectangle has area \((1/4)(5/8)=5/32\). This depicts the complete finite inequality used in (5.1), with no sampled fluid solution. Human source comparison: Tao, article.tex 632–647. Reproducible figure source.

5. Selecting an earlier physical frequency

Put \(r_j=N_j^{-1}\). Combining (3.3) and (4.3),

Thus some index \(i\) has \(r_i>\sqrt S/(DB)\geq\sqrt S/\beta\). The original scale condition gives \(r_0\leq\sqrt S/(2\beta)\), so this index cannot be zero. Take \(1\leq i\leq m\). Then \(N_i<\beta S^{-1/2}\). The incoming step gives the other two bounds:

The non-strict version yields \(N_i\geq\sqrt d\,\eta^3S^{-1/2}\); the strict final comparison implies the required non-strict time separation. Equations (3.1) and (4.3) supply the amplitude and displacement. This proves every assertion of (2.2), with no auxiliary solution-dependent maximum in its constants.

6. Recovering the exact source time separation

We next prove the map to the source's unit-viscosity equation and its actual \(A_{\rm src}\geq2\). This comparison does not replace (1.1)–(5.1). Use exactly the constants \(a_\Delta,\zeta,e_0,c_0,\mathcal L(C),\mathcal D(C)\) of NS-FLUID-15 (7.1)–(7.4). For \(C\geq16\) obeying those selection inequalities, set \(b=A_{\rm src}^{-C}\). Their proved bounds are

All comparisons retain the actual definitions before estimating them. Put \(s_0=s_1+s_2+s_3+s_4>0\); (1.2) gives \(\mathscr S(A_{\rm src},1)\leq s_0A_{\rm src}^4\). Define

These are new comparison constants; \(\mathcal B(C)\) is not the physical \(B\) in (2.1), and \(\mathcal A(C)\) is not the source bound \(A_{\rm src}\).

Since \(1+\eta^{-3}\leq2\eta^{-3}\), substitution of every factor into (2.1) gives

For the first exponent, the four contributions are \(6C+9\) from \(\eta^{-3}\), \(4\) from total speed, \(C\) from \(b^{-1}\), and \(13C+20\) from \(d^{-1}\). They sum to \(20C+33\). The second entry of \(\beta\) is bounded by \(\sqrt{2/a_\Delta}\). The lower product \(d\eta^6\) has coefficient \(e_0\zeta^{-6}\) and exponent \(-25C-38\). Division by \(8D\) gives the stated \(\rho\) bound; division by \(\beta^2\) gives the stated \(\gamma\) bound. This accounts for all original factors.

Choose the smallest integer \(C_\dagger\geq16\) satisfying NS-FLUID-15 (7.4) and all six inequalities

As before \(\log_2^+x=\max\{0,\log_2x\}\). This is a terminating integer selection, not an assumption. Here is a complete growth bound. The proof in NS-FLUID-15 gives \(\mathcal L(C)\leq\mathcal L(1)C^{1/40}\) and \(\mathcal D(C)\leq D_*C\), where

Therefore, with \(B_*=\max\{D_*b_*,\sqrt{2/a_\Delta}\}\) and \(A_*=\max\{2D_*,4B_*^2\}\), we have

The positive logarithm of each right side is at most its constant positive logarithm plus a fixed multiple of \(\log_2 C\). For \(C\geq1\), \(\log_2C\leq2C\). Thus all right sides of (6.4), and those of the prior selection, are bounded by explicit affine functions of \(C\). Their cubes and fourth powers dominate for all sufficiently large integers; the earlier first affine condition is itself eventually true. This proves nonemptiness and finite termination of the selection. One may enlarge the source's unspecified sufficiently large hierarchy exponent to this actual integer.

Use the original source notation \(A_j=A_{\rm src}^{C_\dagger^j}\). For \(A_{\rm src}\geq2\), every fixed positive coefficient \(q\) is bounded above by \(A_{\rm src}^{\log_2^+q}\). Its reciprocal version gives the corresponding lower bounds. Equations (6.3)–(6.4) therefore prove

Suppose the source interval \(A_4N_0^{-2}\leq S\leq A_4^{-1}T\) is nonempty and its original amplitude is \(|P_{N_0}u(t_0,x_0)|\geq A_1^{-1}N_0\). The prior selection ensures \(b=A_1^{-1}\leq b_0\). By (6.6) this is an input to the already proved theorem (2.2). Its actual output satisfies

This proves the original stated time separation, with explicit frequency and displacement bounds. It also proves that the coarser printed \(A_3^{-6}S\) calculation is not a reason to stop at a weaker time conclusion: the preceding full one-step constants have polynomial dependence on \(A_{\rm src}\) with exponents affine in the selected \(C_\dagger\), which fits inside the actual cubic hierarchy. The exact map is (6.1)–(6.7).

7. Every original physical scale

For \(\lambda>0\), keep \(u^\lambda(t,x)=\lambda u(\lambda^2t,\lambda x)\) and \(p^\lambda(t,x)=\lambda^2p(\lambda^2t,\lambda x)\). Direct differentiation retains the same viscosity \(\nu\). The critical bound \(U\) and every constant in (1.2), (2.1) are unchanged. The original physical data map as

The full Fourier substitution gives \(P_{\lambda N}u^\lambda(t,x) =\lambda(P_Nu)(\lambda^2t,\lambda x)\). Every target and earlier amplitude thus has factor \(\lambda\). The receiving time bounds have factor \(\lambda^{-2}\), frequency bounds factor \(\lambda\), and displacement bound factor \(\lambda^{-1}\). Each persistence interval maps with its exact endpoints. Substitution in the total-speed integral gives factor \(\lambda^{-1}\), exactly the factor of \(\sqrt S\). The one-step kernels retain both invariants \(\nu\tau N^2\) and \(N\,\mathrm{distance}\). This proves the full physical comparison without changing the working coordinates of (1.1).

8. Five solved exercises

Exercise 1: use the full persistence interval

For the same actual chain, replace the half-length interval in Section 4 by

Prove the exact amplitude and total-speed bounds. Choose the best permitted \(\theta\), and propagate the resulting constant through every conclusion of the theorem.

Solution. The complete original derivative estimate gives

Since the actual next gap is at least \(dN_j^{-2}\), these intervals remain contained in their separate time steps. They may share endpoints when \(\theta=1\), but those have zero Lebesgue measure. The projection-kernel bound and their exact lengths give

The derivative of \(\theta(1-\theta\eta^6)\) is \(1-2\theta\eta^6>0\) for \(0<\theta\leq1\), since \(\eta\leq1/256\). Thus \(\theta=1\) gives the largest coefficient among these permitted intervals.

Define the actual improved constants

Retain the original \(\rho=d\eta^6/(8D)\). The finite-chain crossing uses only \(\Lambda_{\rm full}\geq2D\) and \(S\leq\rho T\), so Section 3 applies unchanged. Equation (8.2) with \(\theta=1\), followed by the same final frequency comparison, gives the complete sum bound \(\sum_{j=0}^mN_j^{-1}\leq B_{\rm full}\sqrt S\). The full finite sum argument of Section 5 then yields

whenever \(\Lambda_{\rm full}N_0^{-2}\leq S\leq\rho T\). This proves the entire receiving theorem with the new constants. Moreover \(B_{\rm full}<B\), \(\beta_{\rm full}\leq\beta\), \(\Lambda_{\rm full}\leq\Lambda\), and \(\gamma_{\rm full}\geq\gamma\). The lower frequency bound, original amplitude, viscosity and terminal time allowance remain exactly the same. The source comparisons in Section 6 therefore continue to hold, and their full original bounds are retained.

Exercise 2: retain all earlier frequencies selected by the sum

Prove a quantitative statement about the entire set of indices with sufficiently large reciprocal frequency, rather than choosing only one maximum. Keep its actual time, spatial and amplitude information.

Solution. Use the original constants of Section 2 and put

For the complementary indices, \(\sum_{j\notin\mathcal G}r_j^2 \leq a\sum_{j=0}^mr_j\leq S/(2D)\). Subtract this from the complete strict lower bound \(\sum_{j=0}^mr_j^2>S/D\). It gives

Thus \(\mathcal G\) is nonempty. The original scale condition implies \(r_0\leq\sqrt S/(2\beta)\leq a\), so every index in \(\mathcal G\) is at least one. For each of them the incoming gap and frequency comparison give

Indeed \(r_j>a\) makes the actual time gap exceed \(d\eta^6a^2\); all the other conclusions were proved for every \(1\leq j\leq m\). This preserves a positive lower sum of reciprocal-frequency squares over a set of actual earlier points, with all their receiving data. It asserts no lower bound on the number of distinct spatial points: selected points may share a spatial coordinate.

Exercise 3: when does the receiving time interval exist?

Give an exact condition for the interval of permitted \(S\)'s to be nonempty. Deduce the original one-step threshold without assuming it separately. Compare the result for the improved constant in Exercise 1.

Solution. All quantities are positive. The exact equivalence is

No change in the physical \(N_0,T\) is involved. Since \(\Lambda\geq2D\) and \(\rho\leq1/32\), this implies \(N_0^2\geq64D/T\), hence in particular \(N_0\geq2\sqrt{D/T}\), the already proved operation's input condition. The factor 64 is not discarded. When the interval is empty the iterated assertion has no receiving time; it does not imply that the initial amplitude is impossible.

For Exercise 1 the exact condition is \(N_0^2\geq8D\Lambda_{\rm full}/(d\eta^6T)\). It is no stronger than (8.7), and the same initial one-step threshold follows because \(\Lambda_{\rm full}\geq2D\). This proves precisely how the stronger persistence estimate enlarges the admissible range of original frequencies and times.

Exercise 4: reconstruct the source's positive time separation

Starting from the actual original constants, recover the exponent \(89C+142\) and its coefficient in the lower bound for \(\gamma\). Prove why an estimate with that affine exponent can imply the source's cubic-hierarchy separation.

Solution. The full earlier comparison gives

Divide the first expression by the second, retaining both coefficients and all powers:

For \(q>0\) and \(A_{\rm src}\geq2\), \(q\geq A_{\rm src}^{-\log_2^+(1/q)}\). If \(q\geq1\) this follows from \(q\geq1\); if \(0<q<1\), raising \(A_{\rm src}\geq2\) to the negative exponent gives \(A_{\rm src}^{-\log_2(1/q)}\leq q\). Thus the first numerical inequality in (6.4) implies \(\gamma\geq A_{\rm src}^{-C_\dagger^3}=A_3^{-1}\).

This is an actual finite choice: (6.5) gives \(1/\mathcal G(C)\leq K C^2\), with the complete \(K=\zeta^6B_*^2/e_0>0\). Therefore \(\log_2^+(1/\mathcal G(C)) \leq\log_2^+K+2\log_2C\leq\log_2^+K+4C\). The cubic dominates the resulting affine expression \(93C+142+\log_2^+K\). The other simultaneous conditions are proved to be eventual in Section 6, so the finite integer selection satisfies all of them. No unproved replacement of one hierarchy level by another is used.

Exercise 5: a moving frame and the critical class

For a constant vector \(v\in\mathbb R^3\), prove the exact Galilean map of the original equation and the earlier points. Determine whether a nonzero moving frame preserves the whole-space \(L^3\) hypothesis, and retain the displacement term that the moving frame introduces.

Solution. Define the actual fields

The time derivative contributes \(\partial_tu(t,x-vt)-v\cdot\nabla u(t,x-vt)\). The transport term is \((u(t,x-vt)+v)\cdot\nabla u(t,x-vt)\). Their two \(v\)-terms cancel exactly. Pressure gradient, Laplacian and divergence are translated unchanged, so the original unforced equation holds with the same viscosity.

Since the annular symbol vanishes at zero, its projection annihilates the constant vector. Translation commutes with the convolution kernel. Hence

Every selected point maps to \((t_j,x_j^v)=(t_j,x_j+vt_j)\), with unchanged frequency, time and projected amplitude. Its full displacement is

The advective contribution is retained.

If \(v\ne0\), then for each time

Its complement has infinite measure, and there \(|u(t,x-vt)+v|\geq|v|/2\). Consequently \(\|u^v(t)\|_3=\infty\). The transformed solution therefore does not satisfy the original whole-space critical-norm hypothesis. Its fluctuation \(u^v-v\) has the unchanged \(L^3\) norm; in moving coordinates it is exactly the original field. Equations (8.9)–(8.12) prove the relation completely, including the class restriction and actual point map. They do not silently apply the theorem to an infinite critical norm.

9. Source comparison and the next argument

The original proof at article.tex 624 uses \(A_2\) in the first time interval although its preceding step bound uses \(A_3\). Equation (3.1) retains the actual \(d,D\) time constants. At 627 the displayed chain has an ill-ordered comparison and does not imply its stated stopping-time conclusion; (3.2) proves the complete positive bound with its physical \(T\). At 629–630 the receiving center is the original \(t_0\), and the crossing after the largest admissible index has \(t_{m+1}<t_0-S\), as proved in (3.3). Equations (4.1)–(4.3) prove the disjoint interval and total-speed map with all constants. Equation (5.1) uses the actual incoming frequency comparison for the lower time bound. Finally (6.1)–(6.7) prove the stronger source-stated time separation after an explicit fixed hierarchy choice.

These corrections preserve and establish the original iterated conclusion; they do not disprove it. The next argument connects the original vorticity mass, the full Gaussian factors, actual annular supports and shell multiplicities to the finite weighted heat estimates and the small annulus constructed earlier. Those receiving arguments are needed before claiming the general large critical-velocity endpoint. No novelty is asserted here.

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